Consider on $\mathbb{R}^2$ the distribution defined by the locally integrable function $$ E(x,t) = \begin{cases} 1/2 & t - |x| > 0 \\ 0 & t - |x| < 0 .\end{cases} $$
Compute the distributional derivative $\frac{\partial^2 E}{\partial t^2} - \frac{\partial^2 E}{\partial x^2}.$
I'm not sure how to even start this problem, since to me, $E$ looks like a constant valued function whose derivative should be $0$ everywhere.
Denote by $T$ the distribution defined by $E$. Moreover, let $U$ be the domain on which $E\neq 0$. Then for $\phi\in C_0^\infty(\mathbb R^2)$ we have $$ T\phi = \int E(x,y)\phi(x,y)\,d(x,y) = \frac 12\int_U\phi(x,y)\,d(x,y). $$ Now, by definition, \begin{align*} T_{yy}\phi &= T\phi_{yy} = \frac 12\int_U\phi_{yy}(x,y)\,d(x,y) = \frac 12\int_{\mathbb R}\int_{|x|}^\infty\phi_{yy}(x,y)\,dy\,dx\\ &= -\frac 12\int_{\mathbb R}\phi_{y}(x,|x|)\,dx = -\frac 12\int_0^\infty\big[\phi_y(-x,x)+\phi_y(x,x)\big]\,dx. \end{align*} Similarly, \begin{align*} T_{xx}\phi &= T\phi_{xx} = \frac 12\int_U\phi_{xx}(x,y)\,d(x,y) = \frac 12\int_0^\infty\int_{-y}^y\phi_{xx}(x,y)\,dx\,dy\\ &= \frac 12\int_0^\infty\big[\phi_{x}(y,y)-\phi_x(-y,y)\big]\,dx. \end{align*} Hence, $$ T_{yy}\phi-T_{xx}\phi = -\frac 12\int_0^\infty\big[\phi_y(-t,t)+\phi_y(t,t) + \phi_{x}(t,t)-\phi_x(-t,t)\big]\,dt. $$ Now, let $f(t) := (t,t)$. Then $$ (\phi\circ f)'(t) = \phi'(f(t))f'(t) = (\phi_x(t,t),\phi_y(t,t))\binom 11 = \phi_x(t,t) + \phi_y(t,t). $$ Similarly, if we set $g(t) = (-t,t)$, then $(\phi\circ g)'(t) = -\phi_x(-t,t) + \phi_y(-t,t)$. That is, $$ T_{yy}\phi-T_{xx}\phi = -\frac 12\int_0^\infty\big[(\phi\circ f)'(t) + (\phi\circ g)'(t)\big]\,dt = \frac 12\big[\phi(f(0)) + \phi(g(0))\big] = \phi(0,0). $$ Therefore, $T_{yy}-T_{xx} = \delta$.